Php uses foreach to traverse the array cyclically. The execution in the loop is successful, but it is not successful outside the loop? -SegmentFault

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Foreach ($ arr as $ value) {$ value = preg_replace ("/href = \" \ // I ", 'href = "'. $ link, $ value); echo $ value."
"; // Here $ value is the value after replacement. } Echo $ arr [1]; // The value displayed here is still the value of $ arr [1] before replacement. // What is this? ''

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Foreach ($ arr as $ value) {$ value = preg_replace ("/href = \" \ // I ", 'href = "'. $ link, $ value); echo $ value."
"; // Here $ value is the value after replacement. } Echo $ arr [1]; // The value displayed here is still the value of $ arr [1] before replacement. // What is this? ''

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The answer is helpful to people and has reference value.0The answer is not helpful. It is a wrong answer.

Foreach ($ arr as & $ value) {$ value = preg_replace ("/href = \" \ // I ", 'href = "'. $ link, $ value); echo $ value."
"; // Here $ value is the value after replacement .} Echo $ arr [1]; // The value displayed here is still the value of $ arr [1] before replacement.

If you modify the value in the variable scope and save the result, you need to define the typeReference Type, That is&.

Change to this

Foreach ($ arr as $ key => $ value) {$ arr [$ key] = preg_replace ("/href = \" \ // I ", 'href = "'. $ link, $ value); echo $ arr [$ key]."
"; // Here $ value is the value after replacement. }

You just performed a replace operation on the local variable $ value in the loop body, and did not return the value to $ arr.

Foreach ($ arr as $ key => $ value) {$ value = preg_replace ("/href = \" \ // I ", 'href = "'. $ link, $ value); $ arr [$ key] = $ value; echo $ value."
"; // Here $ value is the value after replacement .} Echo $ arr [1]; // The value displayed here is still the value of $ arr [1] before replacement.

First, let's figure out how to pass by value and by reference.